The speed limit near a school is 30km/h but Joe Cool thinks he can ignore it. He is driving at 60km/h. To make matters worse, he is talking on his cell phone. Suddenly a child rides his bicycle out of school grounds and across the road on the pedestrian crossing. The maximum rate at which Joe can slow down is 8m/s squared. How far away from the child must Joe be when he first sees the child? (before he applies the brakes) in order to stop in time.
Thanks
v^2 = 2ax.
You'll have to turn the kph into m/s as well.
To find the distance Joe must be from the child when he first sees them in order to stop in time, we need to consider Joe's initial speed, his maximum deceleration rate, and the speed limit near the school.
Let's start by converting Joe's speed from km/h to m/s:
60 km/h * (1000 m/1 km) * (1/3600 h/1 s) = 16.67 m/s
Next, we'll calculate the time it would take for Joe to come to a stop using his maximum deceleration rate:
Using the equation v^2 = u^2 + 2as, where:
v = final velocity (0 m/s, since Joe needs to come to a stop),
u = initial velocity (16.67 m/s),
a = acceleration (maximum deceleration rate of 8 m/s^2), and
s = distance stopping distance to be determined.
0^2 = (16.67 m/s)^2 + 2 * 8 m/s^2 * s
Simplifying the equation:
0 = 277.78 m^2/s^2 + 16 m/s^2 * s
Solving for s:
-16 m/s^2 * s = 277.78 m^2/s^2
s = -277.78 m^2/s^2 / -16 m/s^2
s = 17.36 m
Therefore, Joe must be at least 17.36 meters away from the child when he first sees them in order to stop in time.